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|---|---|---|---|---|---|---|---|---|---|
29.28. Calculate the volume $V$ of a cone with radius $R$ and height $h$ using a definite integral. | 29.28. Answer: $\frac{1}{3} \pi R^{2} h$. Let's choose the $O x$ axis as the axis of the cone and place the origin of the coordinates at the vertex of the cone. Then $V=\int_{0}^{h} \pi\left(\frac{R}{h} x\right)^{2} d x=\frac{1}{3} \pi R^{2} h$. | \frac{1}{3}\piR^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,967 |
29.29. Calculate the volume $V$ of a sphere of radius $R$ using a definite integral. | 29.29. Answer: $\frac{4}{3} \pi R^{3}$. A plane at a distance $x$ from the center of a sphere cuts a circle of radius $\sqrt{R^{2}-x^{2}}$ on the sphere. The area of this circle is $\pi\left(R^{2}-x^{2}\right)$, therefore
$$
V=\int_{-R}^{R} \pi\left(R^{2}-x^{2}\right) d x=2 \pi R^{3}-\left.\pi \frac{x^{3}}{3}\right|_{... | \frac{4\piR^{3}}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,968 |
29.30. Find the volume of the figure formed by the intersection of two straight circular cylinders of radius $R$, the axes of which are perpendicular and intersect. | 29.30. Answer: $\frac{16}{3} R^{3}$. Let's introduce a coordinate system, directing the axes $O x$ and $O y$ along the axes of the cylinders. The intersection of the cylinders is described by the inequalities $x^{2}+z^{2} \leqslant R$ and $y^{2}+z^{2} \leqslant R$. Its section by the plane $z=z_{0}$ is described by the... | \frac{16}{3}R^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,969 |
29.32. Prove that if the function $f(x)$ on the interval $[a, b]$ has a continuous derivative, then the curve $y=f(x)$ has length $\int_{a}^{b} \sqrt{1+f^{\prime 2}(x)} d x$. | 29.32. It is clear that $A_{k} A_{k+1}=\sqrt{\left(x_{k+1}-x_{k}\right)^{2}+\left(f\left(x_{k+1}\right)-f\left(x_{k}\right)\right)^{2}}$. By the Lagrange theorem,
$$
f\left(x_{k+1}\right)-f\left(x_{k}\right)=f^{\prime}\left(\xi_{k}\right)\left(x_{k+1}-x_{k}\right)
$$
for some point $\xi_{k}$ in the interval $\left[x_... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,971 |
29.33. Calculate the length of a circle of radius $R$ using a definite integral. | 29.33. The length of the circumference is equal to twice the length of the semicircle $y=\sqrt{R^{2}-x^{2}}$. It is clear that $y^{\prime}=\frac{-x}{\sqrt{R^{2}-x^{2}}}$, so the length of the semicircle is
$$
\lim _{\varepsilon \rightarrow 0} \int_{-R+\varepsilon}^{R-\varepsilon} \sqrt{1+y^{\prime 2}} d x=\lim _{\vare... | \piR | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,972 |
29.34. Calculate the length of the arc of the parabola $2 y=x^{2}$ from the point $(0,0)$ to the point $\left(x_{0}, \frac{x_{0}^{2}}{2}\right)$. | 29.34. The length of the considered arc of the parabola is equal to
$$
\int_{0}^{x_{0}} \sqrt{1+y'^2} d x=\int_{0}^{x_{0}} \sqrt{1+x^{2}} d x=\frac{x_{0} \sqrt{1+x_{0}^{2}}}{2}+\frac{1}{2} \ln \left(x_{0}+\sqrt{1+x_{0}^{2}}\right)
$$
(see problem 29.7). | \frac{x_{0}\sqrt{1+x_{0}^{2}}}{2}+\frac{1}{2}\ln(x_{0}+\sqrt{1+x_{0}^{2}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,973 |
29.35. Calculate the length of the arc of the curve $y=\operatorname{ch} x$, enclosed between the points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$. | 29.35. It is clear that $y^{\prime}=\operatorname{sh} x$ and $\sqrt{1+\left(y^{\prime}\right)^{2}}=\operatorname{ch} x$. Therefore, the length of the considered arc is $\int_{x_{1}}^{x_{2}} \operatorname{ch} x d x=\operatorname{sh} x_{2}-\operatorname{sh} x_{1}$ (it is assumed that $\left.x_{1}<x_{2}\right)$.
| \operatorname{sh}x_{2}-\operatorname{sh}x_{1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,974 |
29.36. Let the graph of the function $y=f(x)$ on the interval $[a, b]$ be parameterized by a parameter $t$, i.e., a monotonically increasing function $x(t)$ is given, such that $a=x\left(t_{0}\right)$ and $b=x\left(t_{1}\right)$, and we assume $y(t)=f(x(t))$. Prove that the length of this graph is $\int_{t_{0}}^{t_{1}}... | 29.36. By the formula for change of variables
$$
\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=\int_{t_{0}}^{t_{1}} \sqrt{1+\frac{y^{\prime 2}}{x^{\prime 2}}} x^{\prime} d t=\int_{t_{0}}^{t_{1}} \sqrt{x^{\prime 2}+y^{\prime 2}} d t
$$
in writing the last equality, we used the fact that $x^{\prime}>0$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,975 |
29.37. Calculate the length of the astroid given by the equation $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$. | 29.37. Ans: $6 a$. The astroid can be given parametrically: $x=a \sin ^{3} t, y=a \cos ^{3} t$. It is sufficient to compute the length of a quarter of the astroid, for which $x$ and $y$ are non-negative. Clearly, $x'=$ $=3 a \cos t \sin ^{2} t$ and $y'=-3 a \sin t \cos ^{2} t$. Therefore, $\sqrt{x'^{2}+y'^{2}}=3 a \sin... | 6a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,976 |
29.38. Calculate the length of the cycloid branch
$$
x=a(t-\sin t), \quad y=a(1-\cos t), \quad \text { where } \quad 0 \leqslant t \leqslant 2 \pi
$$
## 29.7. Surface Area
Let $f(x)$ be a continuous positive function on the interval $[a, b], x_{0}=a<x_{1}<\ldots<x_{n-1}<b=x_{n}$ - some points on the interval $[a, b]... | 29.38. Answer: $8a$. It is clear that
$$
\sqrt{x'^2 + y'^2} = a \sqrt{(1 - \cos t)^2 + \sin^2 t} = 2a \sin \frac{t}{2}
$$
Therefore, the length of the branch of the cycloid is
$$
2a \int_{0}^{2\pi} \sin \frac{t}{2} dt = -\left.4a \cos \frac{t}{2}\right|_{0}^{2\pi} = 8a
$$ | 8a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,977 |
29.39. Prove that if a positive function $f(x)$ on the interval $[a, b]$ has a continuous derivative, then the surface area of the surface of revolution generated by the curve $y=f(x)$ is
$$
2 \pi \int_{a}^{b} f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$ | 29.39. We assume it is known from stereometry that the lateral surface area of a truncated cone is equal to the product of the generatrix by the length of the middle section. The lateral surface area of the cone formed by the rotation of the segment $A_{k} A_{k+1}$ is
$$
\pi\left(f\left(x_{k}\right)+f\left(x_{k+1}\rig... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,978 |
29.40. Prove that the surface area of a sphere of radius $R$, enclosed between two parallel planes (intersecting the sphere), is equal to $2 \pi R h$, where $h$ is the distance between these planes.
## 29.8. Inequalities | 29.40. We consider the surface of a sphere as a figure formed by rotating the curve $y=\sqrt{R^{2}-x^{2}}$. Clearly, $y^{\prime}=\frac{-x}{\sqrt{R^{2}-x^{2}}}$ and $\sqrt{1+y^{\prime 2}}=\frac{R}{\sqrt{R^{2}-x^{2}}}$. Therefore, the considered area is
$$
2 \pi \int_{a}^{a+h} \sqrt{R^{2}-x^{2}} \frac{R}{\sqrt{R^{2}-x^{... | 2\piR | Geometry | proof | Yes | Yes | olympiads | false | 38,979 |
29.41. Let $t>0$.
a) Prove that $t-\frac{t^{3}}{6} \leqslant \sin t \leqslant t$ and $1-\frac{t^{2}}{4} \leqslant \cos t \leqslant 1-\frac{t^{2}}{2}+\frac{t^{4}}{24}$.
b) Prove that
$$
\begin{aligned}
& t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\ldots-\frac{t^{4 n+3}}{(4 n+3)!} \leqslant \sin t \leqslant t-\frac{t^{3}}{3!}... | 29.41. From the inequality $\cos x \leqslant 1$, it follows that $\int_{0}^{t} \cos x d x \leqslant \int_{0}^{t} 1 d x$, i.e., $\sin t \leqslant t$. From the obtained inequality, it follows that $\int_{0}^{t} \sin x d x \leqslant$ $\leqslant \int_{0}^{t} x d x=\frac{t^{2}}{2}$, i.e., $1-\cos t \leqslant \frac{t^{2}}{2}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,980 |
29.42. Prove that if $0 < x < y$, then $\cos y < \cos x$.
Prove that if $0 < t < \frac{\pi}{2}$, then $\sin t > t - \frac{t^3}{6}$ and $\cos t < 1 - \frac{t^2}{2}$. | 29.42. Using the inequalities $t-\frac{t^{3}}{6} \leqslant \sin t$ and $\cos t \leqslant 1-\frac{t^{2}}{2} + \frac{t^{4}}{24}$ (Problem 29.41), it follows from the first inequality that
$$
\left(\frac{\sin t}{t}\right)^{3} \geqslant \left(1-\frac{t^{2}}{6}\right)^{3}=1-\frac{t^{2}}{2}+\frac{t^{4}}{12}-\frac{t^{6}}{216... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,981 |
29.43. Prove that if $0<t \leqslant \pi / 2$, then $\frac{1}{\sin ^{2} t} \leqslant \frac{1}{t^{2}}+1-\frac{4}{\pi^{2}}$. | 29.43. Consider the function $f(t)=\frac{1}{\sin ^{2} t}-\frac{1}{t^{2}}$. It is clear that $f\left(\frac{\pi}{2}\right)=$ $=1-\frac{4}{\pi^{2}}$. Moreover,
$$
f^{\prime}(t)=\frac{2}{t^{3}}-\frac{2 \cos t}{\sin ^{3} t}=\frac{2}{\sin ^{3} t}\left(\left(\frac{\sin t}{t}\right)^{3}-\cos t\right)
$$
Therefore, according ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,982 |
29.46. Prove that for any $t>0$
$t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\frac{t^{7}}{7}+\ldots-\frac{t^{4 n-1}}{4 n-1}<\operatorname{arctg} t<t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\frac{t^{7}}{7}+\ldots+\frac{t^{4 n+1}}{4 n+1}$. | 29.46. We will use the fact that $\int_{0}^{t} \frac{d x}{1+x^{2}}=\operatorname{arctg} t$. For $\frac{1}{1+x^{2}}$, we have the inequalities
$$
1-x^{2}+x^{4}-x^{6}+\ldots-t^{4 n-2}<\frac{1}{1+x^{2}}<1-x^{2}+x^{4}-x^{6}+\ldots+t^{4 n}
$$
Integrating these inequalities, we obtain the required result. | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,985 |
29.47. a) Prove that if $0 \leqslant t \leqslant a$, then
$1+t+\frac{t^{2}}{2!}+\frac{t^{3}}{3!}+\ldots+\frac{t^{n}}{n!} \leqslant e^{t} \leqslant 1+t+\frac{t^{2}}{2!}+\frac{t^{3}}{3!}+\ldots+\frac{t^{n-1}}{(n-1)!}+e^{a} \frac{t^{n}}{n!}$.
b) Prove that if $0 \geqslant t \geqslant a$, then for odd $n$ the same inequa... | 29.47. a) Since $e>1$, then $1 \leqslant e^{x} \leqslant e^{a}$ for $0 \leqslant x \leqslant t$. Therefore, $\int_{0}^{t} 1 d x \leqslant \int_{0}^{t} e^{x} d x \leqslant \int_{0}^{t} e^{a} d x$, i.e., $t \leqslant e^{t}-1 \leqslant e^{a} t$. From the inequality $\quad 0$
$1+x \leqslant e^{x} \leqslant 1+e^{a} x$ simil... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,986 |
29.48. The function $f(x)$ is differentiable on the interval $[a, b]$, the length of which is $\pi$. Prove that there is a point $x$ on the interval $[a, b]$ for which $f^{\prime}(x)-(f(x))^{2}<1$.
## 29.9. Calculation of limits | 29.48. Suppose that if $a \leqslant x \leqslant b$, then $f^{\prime}(x)-(f(x))^{2} \geqslant 1$, i.e., $\frac{f^{\prime}(x)}{1+(f(x))^{2}} \geqslant 1$. Consider the function $F(x)=\operatorname{arctg}(f(x))$. Then $F^{\prime}(x)=\frac{f^{\prime}(x)}{1+(f(x))^{2}} \geqslant 1$, since $(\operatorname{arctg} y)^{\prime}=... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,987 |
29.49. Calculate the limit $\lim _{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{n}{n^{2}+k^{2}}$. | 29.49. Answer: $\pi / 4$. The considered sum can be written as $\sum_{k=0}^{n-1} \frac{1}{n} \frac{1}{1+(k / n)^{2}}$. As $n \rightarrow \infty$, this sum tends to $\kappa \int_{0}^{1} \frac{d x}{1+x^{2}}=\frac{\pi}{4}$. | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,988 |
29.50. Calculate the limit $\lim _{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{\sqrt{n^{2}-k^{2}}}{n^{2}}$. | 29.50. Answer: $\pi / 4$. The considered sum can be written as $\sum_{k=0}^{n-1} \frac{1}{n} \sqrt{1-\left(\frac{k}{n}\right)^{2}}$. As $n \rightarrow \infty$, this sum tends to $\kappa \int_{0}^{1} \sqrt{1-x^{2}} d x=\frac{\pi}{4}$. | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,989 |
29.51. Calculate $\lim _{n \rightarrow \infty} \sum_{k=n}^{2 n} \frac{1}{k}$. | 29.51. Answer: $\ln 2$. Note that the considered sum is equal to
$$
\frac{1}{n}\left(1+\frac{1}{1+1 / n}+\frac{1}{1+2 / n}+\ldots+\frac{1}{1+n / n}\right) \rightarrow \int_{1}^{2} \frac{d x}{1+x}=\ln 2
$$
Remark. Another proof can be found in the solution to problem 30.8 a). | \ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,990 |
29.52. Compute $\lim _{n \rightarrow \infty} \sum_{k=0}^{n-1} \frac{1}{\sqrt{n^{2}-k^{2}}}$.
See also problem 25.42.
### 29.10. Identities | 29.52. Answer: $\pi / 2$. The considered sum can be written as $\sum_{k=0}^{n-1} \frac{1}{n} \frac{1}{\sqrt{1-(k / n)^{2}}}$. This sum tends to $\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}=$ $=\left.\arcsin x\right|_{0} ^{1}=\frac{\pi}{2}$. | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,991 |
29.53. Prove the following identity for binomial coefficients:
$$
C_{n}^{1} \frac{1}{1}-C_{n}^{2} \frac{1}{2}+\ldots+(-1)^{n-1} C_{n}^{n} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}
$$ | 29.53. Let's write the identity $1+q+q^{2}+\ldots+q^{n-1}=\frac{1-q^{n}}{1-q}$ for $q=1-t$. As a result, we get the identity $\sum_{k=0}^{n-1}(1-t)^{k}=\frac{1-(1-t)^{n}}{t}$. The required identity is obtained by integrating this identity from 0 to 1. Indeed, $\int_{0}^{1}(1-t)^{k} d t=\int_{0}^{1} x^{k} d x=\frac{1}{k... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,992 |
29.54. a) Compute $\int_{-1}^{1}\left(x^{2}-1\right)^{n} d x$, where $n$ is a natural number, using the formula from problem 29.13.
b) Prove that
$$
\begin{aligned}
\frac{1}{2 n+1}-C_{n}^{1} \frac{1}{2 n-1}+C_{n}^{2} \frac{1}{2 n-3}-C_{n}^{3} \frac{1}{2 n-5}+\ldots+ & (-1)^{n} C_{n}^{n}= \\
& =\frac{(-1)^{n} 2^{2 n}(... | 29.54. a) The integrand can be written in the form \(P(x)(x+1)^{n}\), where \(P(x)=(x-1)^{n}\). Using the formula from problem 29.13, we get
\[
\begin{array}{r}
\int_{-1}^{1}\left(x^{2}-1\right)^{n} d x=\left((x-1)^{n} \frac{(x+1)^{n+1}}{n+1}-n(x-1)^{n-1} \frac{(x+1)^{n+2}}{(n+1)(n+2)}+\ldots\right. \\
\left.\ldots+(-... | \frac{(-1)^{n}2^{2n+1}(n!)^{2}}{(2n+1)!} | Calculus | proof | Yes | Yes | olympiads | false | 38,993 |
29.55. Prove that for any natural $n \geqslant 3$ there exists a polynomial $f(x)$ of degree $n$ with real coefficients and roots $a_{1}<a_{2}<\ldots<a_{n}$, for which
$$
\int_{a_{1}}^{a_{2}}|f(x)| d x=\int_{a_{2}}^{a_{3}}|f(x)| d x=\ldots=\int_{a_{n}}^{a_{n-1}}|f(x)| d x
$$
### 29.12. Improper Integrals
The integra... | 29.55. The polynomial $f(x)$ has no multiple roots, so the equivalent condition is: $\int_{a_{1}}^{a_{2}} f(x) d x=-\int_{a_{2}}^{a_{3}} f(x) d x=\ldots=\int_{a_{3}}^{a_{4}} f(x) d x=$ $=\ldots$ Let $f(x)=F^{\prime}(x)$ for some polynomial $F$. Then we obtain the following condition: $F\left(a_{2}\right)-F\left(a_{1}\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 38,994 |
30.3. Let $|x|<1$. Compute the sum of the series $\sum_{n=1}^{\infty} n x^{n-1}$. | 30.3. Answer: $\frac{1}{(1-x)^{2}}$. According to problem 9.6
$$
\sum_{n=1}^{m+1} n x^{n-1}=\frac{(m+1) x^{m+2}-(m+2) x^{m+1}+1}{(1-x)^{2}}
$$
It is also clear that $\lim _{m \rightarrow \infty} x^{m}=0$. | \frac{1}{(1-x)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,996 |
30.4. Compute the sum of the series \(\sum_{n=1}^{\infty} \frac{1}{n} \sin n x\).
## 30.2. Computation of infinite products | 30.4. Answer: 0 when $x=2 k \pi$ and $\frac{\pi-x}{2}+\left[\frac{x}{2 \pi}\right] \pi$ in other cases.
When $x$ is replaced by $x+2 k \pi$, the considered sum and the proposed answer do not change. Therefore, it is sufficient to consider
## Chapter 30. Series
the case when $0<x<2 \pi$. Recall that
$$
\cos x+\cos 2... | \frac{\pi-x}{2}+[\frac{x}{2\pi}]\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,997 |
30.5. Let \(a_{1}=1\) and \(a_{n}=n\left(a_{n-1}+1\right)\) for \(n \geqslant 2\). Prove that \(\prod_{n=1}^{\infty}\left(1+\frac{1}{a_{n}}\right)=e\).
## 30.3. Harmonic Series
The series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is called the harmonic series. This series is divergent (Problem 30.6). We will use the notati... | 30.5. From the equality $a_{n+1}=(n+1)\left(a_{n}+1\right)$, it follows that $a_{n}+1=\frac{a_{n+1}}{n+1}$. Therefore, $P_{N}=\prod_{n=1}^{N}\left(1+\frac{1}{a_{n}}\right)=\prod_{n=1}^{N} \frac{a_{n+1}}{(n+1) a_{n}}=\frac{a_{N+1}}{(N+1)!}$. Hence, $P_{N+1}=$ $=P_{N} \frac{a_{N+1}+1}{a_{N+1}}=P_{N}+\frac{P_{N}}{a_{N+1}}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,998 |
30.6. Prove that for any natural $n$
$$
\ln (n+1) < H_{n} < 1 + \ln n
$$ | 30.6. According to problem 28.49 b) $\frac{1}{k+1}<\ln (k+1)-\ln k<\frac{1}{k}$ for any natural $k$. Adding such inequalities for $k=1,2, \ldots$ $\ldots, n$, we get
$$
H_{n+1}-1<\ln (n+1)<H_{n} .
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,999 |
30.7. Prove that the series $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots$ diverges. | 30.7. The inequality $\frac{1}{2 n+1}>\frac{1}{2 n}$ shows that if the series $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots$ were to converge, then the series $\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots\right)$ would also converge. But according to problem 30.6, this series diverges. | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,000 |
30.8. a) Let $a_{n}=\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2 n}$. Compute the limit $\lim _{n \rightarrow \infty} a_{n}$.
b) Let $b_{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2 n}$. Prove that $\lim _{n \rightarrow \infty} b_{n}=\ln 2$. | 30.8. a) Answer: $\ln 2$. According to problem 28.49 b) $\ln \frac{k+1}{k}<\frac{1}{k}<\ln \frac{k}{k-1}$. Adding such inequalities for $k=n, n+1, \ldots, 2 n$, we get $\ln \frac{2 n+1}{n}<a_{n}<\ln \frac{2 n}{n-1}$. Therefore, $\lim _{n \rightarrow \infty} a_{n}=\ln 2$.
Remark. Another proof can be found in the solut... | \ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 39,001 |
30.9. a) Prove that
$$
\frac{1}{1 \cdot 2}+\frac{1}{3 \cdot 4}+\frac{1}{5 \cdot 6}+\frac{1}{7 \cdot 8}+\ldots=\ln 2
$$
b) Prove that
$$
\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots=1-\ln 2
$$ | 30.9. a) Note that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. Therefore, the sum of the series from problem 30.8 b) is equal to the sum of the series in question.
b) This follows directly from problem 30.1 and part a). | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,002 |
30.10. Prove that
$$
1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\ldots=\frac{1}{2} \ln 2
$$ | 30.10. It is sufficient to prove that the sum of the first $3 n$ terms tends to $\frac{1}{2} \ln 2$. This sum is
$$
\begin{aligned}
& 1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{2 n-1}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\ldots-\frac{1}{4 n}= \\
& =H_{2 n}-\frac{1}{2} H_{n}-\frac{1}{2} H_{2 n}=\frac{1}{2}\left(H_{2 n-1}... | \frac{1}{2}\ln2 | Algebra | proof | Yes | Yes | olympiads | false | 39,003 |
30.11. Let \( a_{n}=\frac{1}{2 n+1}+\frac{1}{2 n+3}+\frac{1}{2 n+5}+\ldots+\frac{1}{4 n-1} \). Prove that \( \lim _{n \rightarrow \infty} a_{n}=\frac{1}{2} \ln 2 \). | 30.11. It is easy to check that
$$
a_{n}=H_{4 n-1}-H_{2 n}-\frac{1}{2}\left(H_{2 n-1}-H_{n}\right)
$$
According to problem 30.8 a) $H_{4 n-1}-H_{2 n} \rightarrow \ln 2$ and $H_{2 n-1}-H_{n} \rightarrow \ln 2$. | \frac{1}{2}\ln2 | Calculus | proof | Yes | Yes | olympiads | false | 39,004 |
30.12. Prove that
$$
\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{5 \cdot 6 \cdot 7}+\frac{1}{9 \cdot 10 \cdot 11}+\ldots=\frac{1}{4} \ln 2
$$ | 30.12. It is easy to check that the terms of the series have the form
$$
\frac{1}{2}\left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n-1}\right), \quad n=1,2, \ldots
$$
Therefore, twice the sum of the first $n$ terms of the series is $\frac{1}{2 n+1}+$ $+\frac{1}{2 n+2}+\frac{1}{2 n+3}+\ldots+\frac{1}{4 n-1}$. It rem... | \frac{1}{4}\ln2 | Algebra | proof | Yes | Yes | olympiads | false | 39,005 |
30.13. Prove that
$$
\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{3 \cdot 4 \cdot 5}+\frac{1}{5 \cdot 6 \cdot 7}+\ldots=\ln 2-\frac{1}{2}
$$ | 30.13. Note that
$$
\frac{1}{(n-1) n(n+1)}=\frac{1}{2}\left(\frac{1}{n-1}+\frac{1}{n+1}\right)-\frac{1}{n}
$$
Further, the sum of the terms
$$
\frac{1}{2}\left(1+\frac{1}{3}\right)-\frac{1}{2}, \quad \frac{1}{2}\left(\frac{1}{3}+\frac{1}{5}\right)-\frac{1}{4}, \quad \ldots, \quad \frac{1}{2}\left(\frac{1}{2 n-1}+\fr... | \ln2-\frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 39,006 |
30.14. Prove that
$$
\sum_{k=1}^{n} \frac{x^{k}}{k}=H_{n}+\sum_{k=1}^{n} C_{n}^{k} \frac{(x-1)^{k}}{k}
$$
388 | 30.14. Let $f(x)=\sum_{k=1}^{n}\left(\frac{x^{k}}{k}-C_{n}^{k} \frac{(x-1)^{k}}{k}\right)$. It is clear that $f(1)=H_{n}$. Therefore, it is sufficient to prove that $f^{\prime}(x)=0$ for all $x$. It is clear that $f^{\prime}(x)=\sum_{k=1}^{n} x^{k-1}-\sum_{k=1}^{n} C_{n}^{k}(x-1)^{k-1}$. Let $g(x)=\sum_{k=1}^{n} C_{n}^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,007 |
30.15. Let $p_{1}=2, p_{2}=3, \ldots$ be the sequence of prime numbers.
a) Prove that
$$
\left(1-\frac{1}{2}\right)^{-1}\left(1-\frac{1}{3}\right)^{-1} \ldots\left(1-\frac{1}{p_{m}}\right)^{-1}>\sum_{n=1}^{p_{m}} \frac{1}{n}
$$
b) Prove that $1+\sum_{n=1}^{m} \frac{1}{p_{n}}>\ln \ln p_{m}$. | 30.15. a) It is clear that $\left(1-\frac{1}{p}\right)^{-1}=1+\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}+\ldots$ Therefore, the considered product is the sum of numbers $1 / n$, where $n$ is divisible by at least one of the numbers $2,3, \ldots, p_{m}$ (or $n=1$). All numbers $1 / n$, where $1 \leqslant n \leqslant p_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,008 |
30.16. Prove that the limit exists
$$
\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\ldots+\frac{1}{n}-\ln n\right)
$$
- The limit $\gamma=0.5772157 \ldots$ in problem 30.16 is called the Euler's constant. | 30.16. Consider the sequence
$$
u_{m}=\frac{1}{m}-\ln \frac{m+1}{m}=\int_{0}^{1}\left(\frac{1}{m}-\frac{1}{m+t}\right) d t
$$
It is clear that if $0 \leqslant t \leqslant 1$, then
$$
0 \leqslant \frac{1}{m}-\frac{1}{m+t}=\frac{t}{m(m+t)} \leqslant \frac{1}{m^{2}}
$$
therefore $0 \leqslant u_{m} \leqslant 1 / m^{2}$... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,009 |
30.17. Prove that $\prod_{n=1}^{\infty} \frac{e^{1 / n}}{1+1 / n}=e^{\gamma}$, where $\gamma$ is the Euler's constant.
## 30.4. Series for the logarithm | 30.17. The product of the first $n$ terms is equal to
$$
\frac{e^{1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}}}{n+1}=\frac{n}{n+1} e^{1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\ln n}
$$
It remains to note that $\lim _{n \rightarrow \infty} \frac{n}{n+1}=1, \lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\frac{1}{... | e^{\gamma} | Calculus | proof | Yes | Yes | olympiads | false | 39,010 |
30.18. Prove that if $-1<x<1$, then
$$
\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\ldots
$$ | 30.18. Identity
$$
\frac{1}{1+t}=1-t+t^{2}-\ldots+(-1)^{n-1} t^{n-1}+\frac{(-1)^{n} t^{n}}{1+t}
$$
shows that if $x>-1$, then
$$
\ln (1+x)=\int_{0}^{x} \frac{d t}{1+t}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\ldots+(-1)^{n-1} \frac{x^{n}}{n}+(-1)^{n} R_{n}
$$
where $R_{n}=\int_{0}^{x} \frac{t^{n}}{1+t} d t$
If $0 \leqsl... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,011 |
30.19. Prove that if $-1<x<1$, then
$$
\ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\frac{x^{7}}{7}+\ldots\right)
$$ | 30.19. Subtracting term by term from the series
$$
x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\ldots=\ln (1+x)
$$
the series
$$
-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\frac{x^{5}}{5}-\ldots=\ln (1-x)
$$
we obtain the required result. | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,012 |
30.20. a) Prove that for any natural $n$
$\ln (n+1)=\ln n+2\left(\frac{1}{2 n+1}+\frac{1}{3(2 n+1)^{3}}+\frac{1}{5(2 n+1)^{5}}+\ldots\right)$.
b) Prove that
$$
\ln 2=\frac{2}{3}\left(1+\frac{1}{3 \cdot 9}+\frac{1}{5 \cdot 9^{2}}+\frac{1}{7 \cdot 9^{3}}+\ldots\right)
$$ | 30.20. a) Consider the series from problem 30.19 for $x=\frac{1}{2 n+1}$. Since $\frac{1+x}{1-x}=\frac{n+1}{n}$, we obtain the required result.
b) This follows directly from a) when $n=1$, since $\ln 1=0$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,013 |
30.21. a) Prove that
$$
1<\left(n+\frac{1}{2}\right) \ln \left(1+\frac{1}{n}\right)<1+\frac{1}{12 n(n+1)}
$$
b) Prove that the sequence $a_{n}=\frac{n!e^{n}}{n^{n+1 / 2}}$ has a (finite) limit $a$.
c) Prove that $n!=a n^{n+1 / 2} e^{-n+\frac{\vartheta}{12 n}}$ for some $\vartheta$ lying between 0 and 1. | 30.21. a) The formula from problem 30.20 a) can be rewritten as
$$
\left(n+\frac{1}{2}\right) \ln \left(1+\frac{1}{n}\right)=1+\frac{1}{3(2 n+1)^{2}}+\frac{1}{5(2 n+1)^{4}}+\ldots
$$
It remains to note that the sum of this series is less than
$$
1+\frac{1}{3}\left(\frac{1}{(2 n+1)^{2}}+\frac{1}{(2 n+1)^{4}}+\ldots\r... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,014 |
30.22. Prove that the number $a$ from problem 30.21 is equal to $\sqrt{2 \pi}$, i.e.
$$
\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}<n!<\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n} e^{1 /(12 n)}
$$
(Stirling's formula).
## 30.5. Series for the number $\pi$ | 30.22. The formula of Wallis (problem 29.23) can be rewritten as follows: $\sqrt{\pi}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}} \frac{(2 n)!!}{(2 n-1)!!}$. Further,
$$
\frac{(2 n)!!}{(2 n-1)!!}=\frac{2 \cdot 4 \cdot \cdots \cdot 2 n}{1 \cdot 3 \cdot \ldots \cdot(2 n-1)}=\frac{(2 \cdot 4 \cdot \ldots \cdot 2 n)^{... | \sqrt{2\pi} | Inequalities | proof | Yes | Yes | olympiads | false | 39,015 |
30.23. Let $a_{n}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots+(-1)^{n} \frac{1}{2 n+1}$.
a) Prove that $\lim _{n \rightarrow \infty} a_{n}=\pi / 4$.
b) Prove that $\frac{1}{2(4 k+1)} \leqslant \frac{\pi}{4}-a_{2 k-1} \leqslant \frac{1}{4 k+1}$. | 30.23. It is clear that $\lim _{n \rightarrow \infty}\left(a_{2 k-1}-a_{2 k}\right)=0$, so it is sufficient to solve problem b). From the identity $\frac{1-t^{4 k}}{1+t^{2}}=1-t^{2}+t^{4}-t^{6}+\ldots-t^{4 k-2}$, it follows that
$$
\begin{aligned}
& \frac{\pi}{4}=\operatorname{arctg} 1=\int_{0}^{1} \frac{1}{1+t^{2}} d... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,016 |
30.24. Prove that
$$
1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots=\frac{\pi^{2}}{6}
$$ | 30.24. According to problem $11.1 \operatorname{ctg}^{2} \alpha<\frac{1}{\alpha^{2}}<\frac{1}{\sin ^{2} \alpha}$ for $0<\alpha<\pi / 2$. Summing such inequalities for $\alpha=\frac{\pi}{2 n+1}, \frac{2 \pi}{2 n+1}, \ldots, \frac{n \pi}{2 n+1}$. Using the result of problem 23.7 afterwards, we get
$$
\frac{n(2 n-1)}{3}<... | proof | Calculus | proof | Yes | Yes | olympiads | false | 39,017 |
30.25. a) Prove that
$$
\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2}<\frac{\pi^{2}}{6}\left(a_{1}^{2}+2^{2} a_{2}^{2}+\ldots+n^{2} a_{n}^{2}\right)
$$
b) Prove that
$$
\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{4}<\pi^{2}\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\right)\left(a_{1}^{2}+2^{2} a_{2}^{2}+\ldots+n^{2} a_{n}^{2}... | 30.25. a) According to the Cauchy inequality (problem 1.9)
\[
\begin{aligned}
\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2}=\left(a_{1} & +2 a_{2} \frac{1}{2}+\ldots+n a_{n} \frac{1}{n}\right)^{2} \leqslant \\
& \leqslant\left(1+\frac{1}{2^{2}}+\ldots+\frac{1}{n^{2}}\right)\left(a_{1}^{2}+2^{2} a_{2}^{2}+\ldots+n^{2} a_{n... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,018 |
30.26. Prove that this series converges for any $z$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 30.26. Let $s_{n}=\sum_{k=0}^{n} \frac{z^{k}}{k!}$ and $m<n$. It is clear that $\left|s_{n}-s_{m}\right| \leqslant \sum_{k=m+1}^{n} \frac{|z|^{k}}{k!}$. The series $1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots$, where $x=|z|$, converges. Therefore, using the Cauchy criterion, we obtain that the original series also con... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,019 | |
30.27. Prove that $e^{z} e^{w}=e^{z+w}$ for any complex $z$ and $w$. | 30.27. Let $f_{n}(z)=\sum_{k=0}^{n} \frac{z^{k}}{k!} \cdot$ The equality
$$
\sum_{k+l=m} \frac{z^{k}}{k!} \cdot \frac{w^{l}}{l!}=\frac{1}{m!} \sum_{k+l=m} C_{m}^{k} z^{k} w^{l}=\frac{(z+w)^{m}}{m!}
$$
shows that the expression for $f_{n}(z) f_{n}(w)$ contains all the terms from the expression for $f_{n}(z+w)$ and som... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,020 |
30.28. Prove that if $x$ is a real number, then $e^{i x}=\cos x+i \sin x$ (Euler). | 30.28. By definition
\[
\begin{aligned}
e^{i x} & =1+i x-\frac{x^{2}}{2!}-i \frac{x^{3}}{3!}+\frac{x^{4}}{4!}+i \frac{x^{5}}{5!}-\frac{x^{6}}{6!}-i \frac{x^{7}}{7!}+\ldots= \\
& =\left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\ldots\right)+i\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\ldots\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,021 |
30.30. Prove that for $0<x<\pi / 4$ the inequality $(\sin x)^{\sin x}<(\cos x)^{\cos x}$ holds.
## 30.8. Convergent and Divergent Series | 30.30. The non-triviality of proving this inequality is related to the fact that the function $f(t)=t^{t}$ is not monotonic on the interval $(0,1)$; it has a minimum at the point $t_{0}=1 / e \approx 0.367879$.
Let $u=\sin ^{2} x$ and $v=\cos ^{2} x$. Then $u+v=1$ and $0<\sqrt{u v}<1 / 2$. We need to prove that
$$
\s... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,023 |
30.31. a) Let $a_{1}, a_{2}, a_{3}, \ldots$ be an increasing sequence of natural numbers, in the decimal representation of which the digit 1 does not appear. Prove that the series $\sum_{n=1}^{\infty} 1 / a_{n}$
converges.
b) Let $a_{1}, a_{2}, a_{3}, \ldots$ be an increasing sequence of natural numbers, in the decima... | 30.31. a) If $a_{n} \leqslant 10^{k}-1$, then $n \leqslant 9^{k}-1$, since on these $k$ positions any of the nine digits can stand, and all digits cannot be zeros simultaneously. Therefore, if $n \geqslant 9^{k}$, then $a_{n} \geqslant 10^{k}-1$ and $a_{n+1} \geqslant 10^{k}$. Hence, $\sum_{n=9^{k}+1}^{9^{k+1}} \frac{1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,024 |
30.32. a) Let $a_{1}, a_{2}, a_{3}, \ldots$ be an increasing sequence of natural numbers, and suppose the series $\sum_{n=1}^{\infty} 1 / a_{n}$ diverges. Prove that the number $0, a_{1} a_{2} a_{3} \ldots$ is irrational.
b) Prove that the number $0,12357111317 \ldots$ (consecutive prime numbers are written in sequenc... | 30.32. a) The same reasoning as in the solution to problem 30.31 b) shows that any sequence of digits occurs in one of the numbers $a_{n}$ (the proof needs to be slightly modified for a sequence of all zeros, but we can do without this sequence). From this, it easily follows that the considered decimal fraction is non-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,025 |
30.33. Prove that if $a_{k} \geqslant 0$, then the infinite product $\prod\left(1+a_{k}\right)$ converges if and only if the infinite series $\sum a_{k}$ converges. | 30.33. The sequence $p_{n}=\prod_{k=1}^{n}\left(1+a_{k}\right)$ is non-decreasing, so it either converges to a finite positive number or to $+\infty$. It is not difficult to prove that
$$
a_{1}+\ldots+a_{n} \leqslant\left(1+a_{1}\right) \ldots\left(1+a_{n}\right) \leqslant e^{a_{1}+\ldots+a_{n}}
$$
The first inequali... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,026 |
30.34. Prove that if $0 \leqslant b_{k}<1$, then the infinite product $\prod\left(1-b_{k}\right)$ converges if and only if the infinite series $\sum b_{k}$ converges.
An infinite product $\prod\left(1+a_{k}\right)$ is called absolutely convergent if the infinite product $\prod\left(1+\left|a_{k}\right|\right)$ converg... | 30.34. The sequence $p_{n}=\prod_{k=1}^{n}\left(1-b_{k}\right)$ is non-increasing, so it converges either to a positive number or to zero. It is easy to check that if $0 \leqslant b_{k} < 1$ for all $k \geq N$, then
$$
\frac{p_{n}}{p_{N-1}}=\left(1-b_{N}\right)\left(1-b_{N+1}\right) \ldots\left(1-b_{n}\right) \geqslan... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,027 |
31.1. Prove that if $p-$ is a prime number and $a$ is not divisible by $p$, then
$$
a^{p-1} \equiv 1(\bmod p)
$$
(Fermat's Little Theorem). | 31.1. First solution. According to problem 4.62, the set of remainders when dividing $a, 2a, \ldots, (p-1)a$ by $p$ coincides with the set $1, 2, \ldots, p-1$. Therefore, $a^{p-1} \cdot 1 \cdot 2 \cdot \ldots \cdot (p-1) \equiv 1 \cdot 2 \cdot \ldots \cdot (p-1) \pmod{p}$. The number $b = 1 \cdot 2 \cdot \ldots \cdot (... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,028 |
31.2. Let $p=4k+3$ be a prime number. Prove that if $a^{2}+b^{2}$ is divisible by $p$, then both numbers $a$ and $b$ are divisible by $p$. | 31.2. Suppose that one of the numbers $a$ and $b$ is not divisible by $p$. Then the other number is also not divisible by $p$. Therefore, according to Fermat's Little Theorem, $a^{p-1} \equiv 1(\bmod p)$ and $b^{p-1} \equiv 1(\bmod p)$. Hence, $a^{p-1}+b^{p-1} \equiv 2(\bmod p)$. On the other hand, the number $a^{p-1}+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,029 |
31.3. Prove that there are infinitely many prime numbers of the form $4k+1$. | 31.3. Suppose that $p_{1}, \ldots, p_{r}$ are all the distinct prime numbers of the form $4 k+1$. Consider the number $\left(2 p_{1} \ldots p_{r}\right)^{2}+1$. It is odd, so all its prime divisors must be of the form $4 k \pm 1$. According to problem 31.2, this number cannot have prime divisors of the form $4 k-1$. It... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,030 |
31.4. a) Let $p$ be a prime number. Prove that if $q$ is a prime divisor of the number $2^{p}-1$, then $q-1$ is divisible by $p$.
b) Provide an example of a prime number $p$ for which the number $2^{p}-1$ is not prime.
## 31.2. Pseudoprime numbers
According to Fermat's Little Theorem, for any prime $p$ the number $2... | 31.4. a) Suppose that $2^{p} \equiv 1(\bmod q)$, where $q$ is a prime number. Clearly, $q \neq 2$. If $q-1$ is not divisible by $p$, then the greatest common divisor of $q-1$ and $p$ is 1, so there exist integers $a$ and $b$ such that $a p+b(q-1)=1$ (see p. 43). According to Fermat's Little Theorem, $2^{q-1} \equiv 1(\... | 2^{11}-1=23\cdot89 | Number Theory | proof | Yes | Yes | olympiads | false | 39,031 |
31.5. Prove that the number $341=11 \cdot 31$ is a pseudoprime. | 31.5. The number $2^{341}-2=2\left(2^{340}-2\right)=2\left(\left(2^{10}\right)^{34}-1\right)$ is divisible by $2^{10}-1=1023$. Further, $1023=3 \cdot 341$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,032 |
31.6. Prove that if $n$ is a pseudoprime number, then the number $2^{n}-1$ is also pseudoprime.
Remark. There are even pseudoprime numbers. For example, the number 161038 is pseudoprime.
## 31.3. Euler's Totient Function
| Euler's Totient Function $\varphi(n)$ is equal to the number of integers from $1,2, \ldots$ $\... | 31.6. According to the condition $2^{n}-2=n a$ for some natural $a$. Therefore, $2^{2^{n}-1}-2=2\left(2^{2^{n}-2}-1\right)=2\left(2^{\text {na }}-1\right)$. This number is divisible by $2^{n}-1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,033 |
31.7. a) Prove that if $p$ is a prime number, then $\varphi\left(p^{n}\right)=$ $=p^{n}-p^{n-1}$.
b) Prove that if numbers $m$ and $n$ are coprime, then $\varphi(m n)=\varphi(m) \varphi(n)$. | 31.7. a) Among the numbers $1,2, \ldots, p^{n}-1$, $p^{n-1}-1$ numbers are divisible by $p$, namely, the numbers $p, 2 p, \ldots, p\left(p^{n-1}-1\right)$. Therefore, $\varphi\left(p^{n}\right)=p^{n}-1-\left(p^{n-1}-1\right)=p^{n}-p^{n-1}$.
b) The numbers $m$ and $n$ are coprime, so there exist integers $u$ and $v$ su... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,034 |
31.8. Prove that if the number $a$ is coprime with $n$, then $a^{\varphi(n)} \equiv 1(\bmod n)$ (Euler's theorem). | 31.8. Let $a_{1}, a_{2}, \ldots, a_{\varphi(n)}$ be all numbers from 1 to $n-1$ that are coprime with $n$. We associate the number $a_{i}$ with the remainder of the division of the number $a a_{i}$ by $n$. The number $a a_{i}$ is coprime with $n$, so we will again obtain one of the numbers $a_{1}, a_{2}, \ldots, a_{\va... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,035 |
31.9. The numbers $a$ and $n$ are coprime. Prove that for some natural number $m$ the number $a^{m}-1$ is divisible by $n$.
保留源文本的换行和格式，直接输出翻译结果。 | 31.9. According to Euler's theorem (problem 31.8), we can set \( m = \varphi(n) \). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,036 |
31.10. Prove that $n=\sum \varphi(d)$, where the summation is over all numbers $d$ dividing $n$. | 31.10. First solution. Consider the fractions $1 / n, 2 / n$, $3 / n, \ldots, n / n$; their number is $n$. Replace each of these fractions with the corresponding irreducible fraction. In this case, we obtain fractions whose denominators are divisors of the number $n$, and the number of fractions with denominator $d$ is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,037 |
31.11. a) Find the remainder of the division of $171^{2147}$ by 52.
b) Find the remainder of the division of $126^{1020}$ by 138. | 31.11. a) Answer: 7. The numbers 171 and 52 are coprime, so $171^{\varphi(52)} \equiv 1(\bmod 52)$. Further, $\varphi(52)=\varphi(4) \varphi(13)=24$. Therefore, $171^{2147} \equiv 15^{24 \cdot 89+11} \equiv 15^{11} \equiv 7(\bmod 52)$.
b) Answer: 54. The numbers 126 and 138 are not coprime: their GCD is 6. We will use... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,038 |
31.12. Let $a$ and $m$ be coprime numbers, $k$ be the smallest natural number for which $a^{k} \equiv 1(\bmod m)$. Prove that $\varphi(m)$ is divisible by $k$. | 31.12. According to Euler's theorem $a^{\varphi(m)} \equiv 1(\bmod m)$. Divide $\varphi(m)$ by $k$ with a remainder: $\varphi(m)=k q+r$, where $0 \leqslant r<k$. Then $1 \equiv a^{\varphi(m)} \equiv$ $\equiv\left(a^{k}\right)^{q} a^{r} \equiv a^{r}(\bmod m)$. Therefore, $r=0$, since otherwise we would obtain a contradi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,039 |
31.13. Let $a \geqslant 2$ and $n$ be natural numbers. Prove that $\varphi\left(a^{n}-1\right)$ is divisible by $n$.
Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$ be the prime factorization of the natural number $n$. We will call the generalized Euler's function the function $L(n)$, which is ... | 31.13. The numbers $a$ and $a^{n}-1$ are coprime. Therefore, according to Euler's theorem, $a^{\varphi\left(a^{n}-1\right)} \equiv 1\left(\bmod a^{n}-1\right)$. From this, it follows that the number $\varphi\left(a^{n}-1\right)$ is divisible by $n$ (see problem 31.20).
 Prove that $(p-1)!+1$ is divisible by $p$ if and only if the number $p$ is prime (Wilson's theorem).
b) Prove that
$$
((p-1)!)^{2} \equiv \begin{cases}0(\bmod p), & \text { if } p \text { is composite } \\ 1(\bmod p), & \text { if } p \text { is prime }\end{cases}
$$ | 31.15. a) Suppose that the number $p$ is prime and $p>2$ (for $p=2$ the required statement is obvious). Let $a$ be one of the numbers 1, 2, 3, ..., $p-1$. For it, there exists a unique number $\bar{a}$, $1 \leqslant \bar{a} \leqslant p-1$, such that $a \bar{a} \equiv 1(\bmod p)$ (see the solution to problem 4.63). If $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,041 |
31.16. Let $a(n)$ and $b(n)$ be the remainders when the numbers $((n-1)!)^{2}$ and $((n-1)!+1)^{2}$ are divided by $n$. Prove that $f(n)=n a(n)+2 b(n)$ is a prime number, and that any prime number can be represented in this form.
## 31.5. Problems on Congruences | 31.16. If the number $n$ is prime, then according to Wilson's theorem (problem 31.15 a) $a(n)=1$ and $b(n)=0$, so $f(n)=n$. If, however, the number $n$ is composite, then, as follows from the solution to problem 31.15 b),
$((n-1)!)^{2} \equiv 0(\bmod n)$ and $2(n-1)!\equiv 0(\bmod n)$. Therefore, $a(n)=0$ and $b(n)=1$,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,042 |
31.17. Prove that if the congruences $a \equiv$ $\equiv b\left(\bmod n_{1}\right), \ldots, a \equiv b\left(\bmod n_{k}\right)$ hold, then $a \equiv b(\bmod n)$, where $n=$ $=\operatorname{LCM}\left(n_{1}, \ldots, n_{k}\right)$. | 31.17. The number $a-b$ is divisible by $n_{1}, n_{2}, \ldots, n_{k}$, so it is also divisible by the least common multiple of these numbers. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,043 |
31.18. Let $p$ be a prime number, and $n$ and $a$ be natural numbers not divisible by $p$. Prove that if the congruence $x^{n} \equiv a(\bmod p)$ has a solution, then the congruence $x^{n} \equiv a\left(\bmod p^{r}\right)$ has a solution for any natural number $r$. | 31.18. We apply induction on $r$. Suppose that $x_{r}^{n}=a+m p^{r}$. We will look for an integer $t$ such that for the number $x_{r+1}=x_{r}+t p^{r}$, the congruence $x_{r+1}^{n} \equiv a\left(\bmod p^{r+1}\right)$ holds. Clearly,
$$
\begin{aligned}
& x_{r+1}^{n}=\left(x_{r}+t p^{r}\right)^{n}=x_{r}^{n}+n x_{r}^{n-1}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,044 |
31.19. Prove that if $a \equiv b\left(\bmod p^{n}\right)$, where $p-$ is a prime number, then $a^{p} \equiv b^{p}\left(\bmod p^{n+1}\right)$. | 31.19. By the condition $a=b+m p^{n}$, where $m$ is an integer. Therefore
$$
a^{p}=b^{p}+p b^{p-1} m p^{n}+\frac{p(p-1)}{2} b^{p-2}\left(m p^{n}\right)^{2}+\ldots
$$
All the terms $p b^{p-1} m p^{n}, \frac{p(p-1)}{2} b^{p-2}\left(m p^{n}\right)^{2}, \ldots$ are divisible by $p^{n+1}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,045 |
31.20. Let $a \geqslant 2$ and $n$ be natural numbers. Prove that $a^{k} \equiv 1\left(\bmod a^{n}-1\right)$ if and only if $k$ is divisible by $n$.
$$
\% * *
$$ | 31.20. Let's write $k$ in the form $k=p n+r$, where $0 \leqslant r0$, then $a \leqslant a^{r} \leqslant a^{n-1}<a^{n}-1$. Therefore, $a^{r} \equiv 1\left(\bmod a^{n}-1\right)$ if and only if $r=0$, i.e., the number $k$ is divisible by $n$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,046 |
31.21. Let $p$ be a prime number, $f\left(x_{1}, \ldots, x_{n}\right)$ be a polynomial with integer coefficients, the degree of which in each variable is less than $p$. Prove that if for all integers $x_{1}, \ldots, x_{n}$ the congruence $f\left(x_{1}, \ldots, x_{n}\right) \equiv$ $\equiv 0(\bmod p)$ holds, then all co... | 31.21. We apply induction on $n$. For $n=1$, the statement is true. Indeed, if not all coefficients of the polynomial $f(x)$ are divisible by $p$, then $c f(x) \equiv x^{m} + a_{1} x^{m-1} + \ldots + a_{m} (\bmod p)$ for some integer $c$; here $m < p$. Therefore, according to Lagrange's theorem (problem 31.33), the con... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,047 |
31.22. Let $p$ be a prime number, $f\left(x_{1}, \ldots, x_{n}\right)$ - a polynomial with integer coefficients, for which the congruence $f\left(x_{1}, \ldots, x_{n}\right) \equiv 0(\bmod p)$ holds for all integers $x_{1}, \ldots, \ldots, x_{n}$. Replace in each monomial of this polynomial $x_{i}^{m}$, where $m \geqsl... | 31.22. According to Fermat's Little Theorem, \( x_{i}^{p} \equiv x_{i}(\bmod p) \), therefore \( x_{i}^{m} \equiv x_{i}^{r}(\bmod p) \). Thus, after the specified substitutions, we obtain a polynomial \( g\left(x_{1}, \ldots, x_{n}\right) \), the degree of which in each variable is strictly less than \( p \), and \( g\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,048 |
31.23. Let $f\left(x_{1}, \ldots, x_{n}\right)$ be a polynomial with integer coefficients and a free term equal to zero. Do-
prove that if the degree of this polynomial is less than $n$, then the congruence $f\left(x_{1}, \ldots, x_{n}\right) \equiv 0(\bmod p)$, where $p-$ is a prime number, has a solution different fr... | 31.23. Suppose that the congruence \( f\left(x_{1}, \ldots, x_{n}\right) \equiv 0(\bmod p) \) has only the solution \((0, \ldots, 0)\). Then the congruence
\[
1-\left(f\left(x_{1}, \ldots, x_{n}\right)\right)^{p-1} \equiv\left(1-x_{1}^{p-1}\right) \ldots\left(1-x_{n}^{p-1}\right)(\bmod p)
\]
holds identically. For \(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,049 |
31.24. Let $a, b, c$ be integers. Prove that the congruence $a x^{2}+b y^{2}+c z^{2} \equiv 0(\bmod p)$ has a solution $(x, y, z) \neq$ $\neq(0,0,0)$ for any prime number $p$.
## 31.6. Function $\sigma_{k}(n)$. Divisors
- Let $d_{1}, d_{2}, \ldots$ be the distinct divisors of the number $n$, including 1 and $n$. The ... | 31.24. This statement obviously follows from Chevalley's theorem (problem 31.23), since the degree of the polynomial in question is strictly less than the number of its variables. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,050 |
31.25. a) Prove that if $p$ is a prime number, then
$$
\sigma_{k}\left(p^{a}\right)=1+p^{k}+p^{2 k}+\ldots+p^{a k}
$$
b) Prove that if numbers $m$ and $n$ are coprime, then $\sigma_{k}(m n)=\sigma_{k}(m) \sigma_{k}(n)$.
| A number $n$ is called perfect if $\sigma(n)=2 n$, i.e., the sum of all divisors of the number ... | 31.25. a) The divisors of the number $p^{a}$ are the numbers $1, p, p^{2}, \ldots, p^{a}$.
b) Let $d$ and $d^{\prime}$ be divisors of the numbers $m$ and $n$. Then $d d^{\prime}$ is a divisor of the number $m n$. For coprime numbers $m$ and $n$, the converse is also true: any divisor of the number $m n$ can be uniquel... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,051 |
31.26. a) Prove that if the number $2^{p}-1$ is prime, then the number $2^{p-1}\left(2^{p}-1\right)$ is perfect (Euclid).
b) Prove that if $n$ is an even perfect number, then there exists a prime number of the form $2^{p}-1$ for which $n=2^{p-1}\left(2^{p}-1\right)$ (Euler). | 31.26. a) The numbers $2^{p-1}$ and $\left(2^{p}-1\right)$ are coprime, so according to problem 31.25 b) $\sigma\left(2^{p-1}\left(2^{p}-1\right)\right)=\sigma\left(2^{p-1}\right) \sigma\left(2^{p}-1\right)$. Further, according to problem 31.25 a) $\sigma\left(2^{p-1}\right)=1+2+4+\ldots+2^{p-1}=2^{p}-1$ and $\sigma\le... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,052 |
31.27. Prove that if $n$ is an odd number with exactly two distinct prime divisors, then $\sigma(n)<2 n$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 31.27. Let $n=p_{1}^{a_{1}} p_{2}^{a_{2}}$, where $p_{1}$ and $p_{2}$ are prime numbers, and $p_{1} \geqslant 3$ and $p_{2} \geqslant 5$. Then from the inequality
$$
\sigma(n)=\frac{p_{1}^{a_{1}+1}-1}{p_{1}-1} \cdot \frac{p_{2}^{a_{2}+1}-1}{p_{2}-1}<\frac{p_{1}^{a_{1}+1}}{p_{1}-1} \cdot \frac{p_{2}^{a_{2}+1}}{p_{2}-1}... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,053 | |
31.28. Prove that a natural number $n$ can be chosen such that the ratio $\sigma(n) / n$ can be arbitrarily large. | 31.28. The number $\sigma(n) / n$ is the sum of all numbers of the form $d / n$, where $d$ is a divisor of the number $n$. But $d / n = d^{\prime}$, where $d^{\prime}$ is some divisor of the number $n$, and when $d$ runs through all divisors of the number $n$, $d^{\prime}$ also runs through all divisors of the number $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,054 |
31.29. a) Provide an example of numbers $m \neq n$ for which $\varphi(n)=\varphi(m)$.
b) Prove that if $m \neq n$, then $n \varphi(n) \neq m \varphi(m)$.
c) Provide an example of numbers $m \neq n$ for which $n \sigma(n) = m \sigma(m)$. | 31.29. a) Answer: 5 and 8.
b) Let $n=p_{1}^{a_{1}} \ldots p_{k}^{a_{k}}$ and $p_{1}>p_{2}>\ldots>p_{k}$. Then
$$
n \varphi(n)=\prod_{i=1}^{k}\left(p_{i}^{2 a_{i}-1}\left(p_{i}-1\right)\right)
$$
and the largest prime number that divides $n \varphi(n)$ is $p_{1}$. The maximum power of the number $p_{1}$ that divides ... | 1214 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,055 |
31.30. Prove that
$$
\sum_{k=1}^{2 n} \sigma_{0}(k)-\sum_{k=1}^{n}\left[\frac{2 n}{k}\right]=n
$$
The Möbius function is defined as follows:
$$
\mu(n)= \begin{cases}1 & \text { if } n=1 \\ (-1)^{k} & \text { if } n=p_{1} \ldots p_{k} \\ 0 & \text { if } n=p^{2} m\end{cases}
$$ | 31.30. It is clear that $\left[\frac{2 n+2}{k}\right]-\left[\frac{2 n}{k}\right]=0$ if the numbers $2 n+1$ and $2 n+2$ are not divisible by $k$. Otherwise, this difference is 1. Therefore,
$$
\sum_{k=1}^{n+2}\left[\frac{2 n+2}{k}\right]-\sum_{k=1}^{n}\left[\frac{2 n}{k}\right]=\sigma_{0}(2 n+1)+\sigma_{0}(2 n+2)+1
$$
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,056 |
31.31. Prove that if $F(n)=\sum_{d \mid n} f(d)$, then
$$
f(n)=\sum_{d \mid n} \mu(d) F(n / d)=\sum_{d \mid n} \mu(n / d) F(d)
$$
where the sum is taken over all divisors $d$ of $n$ (Möbius). | 31.31. First, let's check that for all \( n > 1 \), the relation \(\sum_{d \mid n} \mu(d) = 0\) holds. Let \( n = p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}} \). Then
\[
\sum_{d \mid n} \mu(d) = \sum_{d \mid p_{1} \ldots p_{k}} \mu(d) = 1 - C_{k}^{1} + C_{k}^{2} + \ldots + (-1)^{k} = (1-1)^{k} = 0
\]
It is clear tha... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,057 |
31.33. Let $p-$ be a prime number, $f(x)=x^{n}+a_{1} x^{n-1}+\ldots$ $\ldots+a_{n}$ - a polynomial with integer coefficients. Prove that there are no more than $n$ distinct integers $x_{i}$, for which $0 \leqslant x_{i} \leqslant p-1$ and $f\left(x_{i}\right)$ is divisible by $p$ (Lagrange).
An integer $a$ is called a... | 31.33. We apply induction on $n$. For $n=1$, the statement is obvious. Suppose the numbers $f\left(x_{1}\right)$ and $f(x)$ are divisible by $p$ (where $x$ is an integer). Then the number $f(x)-f\left(x_{1}\right)=x^{n}-x_{1}^{n}+a_{1}\left(x^{n-1}-x_{1}^{n-1}\right)+a_{n-1}\left(x-x_{1}\right)=$ $=\left(x-x_{1}\right)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,059 |
31.34. Let $p>2$ be a prime number. Prove that among the numbers $1,2, \ldots, p-1$ exactly half are quadratic residues and exactly half are quadratic non-residues modulo $p$. | 31.34. We associate with each integer $x$, where $1 \leqslant x \leqslant p-1$, the remainder of the division of $x^{2}$ by $p$. The numbers $x$ and $p-x$ are associated with the same number, and $x \neq p-x$. Moreover, according to Lagrange's theorem (problem 31.33$)$, the congruence $x^{2} \equiv c(\bmod p)$ cannot h... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,060 |
31.35. Let $1 \leqslant a \leqslant p-1$, where $p>2$ is a prime number. Prove that the number $a$ is a quadratic residue modulo $p$ if and only if $a^{(p-1) / 2} \equiv 1(\bmod p)$ (Euler). | 31.35. If $a \equiv b^{2}(\bmod p)$, then according to Fermat's Little Theorem (problem 31.1) $a^{(p-1) / 2} \equiv b^{p-1} \equiv 1(\bmod p)$. Therefore, any quadratic residue is a solution to the equation $x^{(p-1) / 2} \equiv 1(\bmod p)$. According to Lagrange's Theorem (problem 31.33), the number of solutions to th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,061 |
31.36. Let $p>2$ be a prime number. Prove that the number -1 is a quadratic residue modulo $p$ if and only if $p \equiv 1(\bmod 4)$. | 31.36. Let's use Euler's criterion (problem 31.35). If $p=4 k+1$, then $(-1)^{(p-1) / 2}=(-1)^{2 k}=1$, and if $p=4 k+3$, then $(-1)^{(p-1) / 2}=(-1)^{2 k+1}=-1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,062 |
31.37. Let $r$ be 2 or an odd number, $p_{1}, \ldots, p_{r}$ be distinct prime numbers of the form $4k+1$. Suppose that $\left(\frac{p_{i}}{p_{j}}\right)=-1$ for all $i \neq j$. Prove that the equation $x^{2}-d y^{2} = -1$, where $d=p_{1} \ldots p_{r}$, has a solution in integers.
## 31.8. Quadratic Reciprocity Law
F... | 31.37. Suppose the equation $x^{2}-d y^{2}=-1$ has no solutions in integers. Let $(\xi, \eta)$ be the fundamental solution of the Pell's equation $x^{2}-d y^{2}=1$. The number $d$ is of the form $4k+1$, so $\xi^{2} \equiv (\eta^{2}+1) \pmod{4}$. The square of an odd number gives a remainder of 1 when divided by 4, so $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,063 |
31.38. Let $p$ be an odd prime number, and $q$ be a natural number not divisible by $p$. For each natural number $l$ from 1 to $\frac{p-1}{2}$, write $l q \equiv \pm r_{l}(\bmod p)$, where $1 \leqslant r_{l} \leqslant \frac{p-1}{2}$. Let $\mu$ be the number of all minus signs encountered here. Prove that $\left(\frac{q... | 31.38. First, note that if $l$ and $k$ are different natural numbers from 1 to $\frac{p-1}{2}$, then $r_{l} \neq r_{k}$. Indeed, if $r_{l}=r_{k}$, then $(l \pm k) q$ is divisible by $p$, so $l \pm k$ is divisible by $p$. But $|l \pm k| \leqslant p-1$.
Thus, the set of numbers $r_{1}, r_{2}, \ldots, r_{(p-1) / 2}$ coin... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,064 |
31.39. Let $p$ be an odd prime number, $a$ a number coprime with $p$, and $\pi_{a, p}: i \mapsto a i(\bmod p)$ a permutation of the residues modulo $p$. Prove that then $\operatorname{sgn} \pi_{a, p}=(a / p)$, where $\operatorname{sgn}=1$ for an even permutation and $\operatorname{sgn}=-1$ for an odd permutation (Zolot... | 31.39. Consider the polynomial $A\left(x_{1}, \ldots, x_{p}\right)=\prod_{1 \leqslant i<j \leqslant p}\left(x_{i}-x_{j}\right)$. Under the action of an even permutation, the polynomial $A$ does not change, while under the action of an odd permutation, it changes sign. Therefore, the sign of any permutation $\sigma$ is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,065 |
31.41. a) Prove the quadratic reciprocity law using problem 31.39.
b) Let $m$ be an odd prime number. Prove that
$$
\left(\frac{2}{m}\right)=(-1)^{\left(m^{2}-1\right) / 8}
$$ | 31.41. a) Let $P=\{0,1, \ldots, p q-1\}$ and $\bar{P}=\{(a, b): 0 \leqslant aq a^{\prime}+b^{\prime}$ and $a+p ba^{\prime}$ and $b<b^{\prime}$. Thus, $k=C_{q}^{2} C_{p}^{2}=\frac{p-1}{2} \cdot \frac{q-1}{2}$. In the end, we get
$$
\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=\operatorname{sgn} \mu \operatorname{sg... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,067 |
31.44. Let $p$ be a prime number. Prove that $\left(\frac{-3}{p}\right)=1$ for $p=6k+1$ and $\left(\frac{-3}{p}\right)=-1$ for $p=6k-1$. | 31.44. It is clear that $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=(-1)^{(p-1) / 2}\left(\frac{3}{p}\right)$. Further, $(-1)^{(p-1) / 2}=1$ for $p=12 k+1$ and $p=12 k+5$, and $(-1)^{(p-1) / 2}=-1$ for $p=12 k-1$ and $p=12 k-5$. Using the result of problem 31.43, we obtain the required ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,070 |
31.45. Prove that there are infinitely many prime numbers of the form $6k+1$. | 31.45. Suppose there are only a finite number of prime numbers of the form $6k+1$. Let $N$ be their product, and $p$ be a prime divisor of the number $4N^2 + 3$. The number $p$ is odd and $p \neq 3$, since $N$ is not divisible by 3. Since $(2N)^2 \equiv -3 \pmod{p}$, we have $\left(\frac{-3}{p}\right) = 1$. According t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,071 |
31.46. a) Let $p=2^{n}-1$ be a prime number, where $p>3$. Prove that $\left(\frac{3}{p}\right)=-1$.
b) Let $p=2^{n}+1$ be a prime number, where $p>3$. Prove that $\left(\frac{3}{p}\right)=-1$. | 31.46. a) It is clear that $p \equiv -1 \pmod{4}$. Moreover, $p \equiv 1 \pmod{3}$. Indeed, $n$ is odd, since the number $2^{2m} - 1$ is divisible by $2^m - 1$, and therefore cannot be prime for $m > 1$. Since $2^2 \equiv 1 \pmod{3}$, it follows that $2^n - 1 \equiv 1 \pmod{3}$ for any odd $n$. As a result, we get that... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,072 |
31.47. Prove that $3^{n}-1$ is not divisible by $2^{n}-1$ for $n>1$.
## 31.9. Gaussian Sums
Let $p-$ be an odd prime number, $\varepsilon=e^{2 \pi i / p}=\cos (2 \pi / p)+$ $+i \sin (2 \pi / p)$. For each integer $a$, we can consider the Gaussian sum $g_{a}=\sum_{k=1}^{p-1}\left(\frac{k}{p}\right) \varepsilon^{a k}$,... | 31.47. If $n=2 m$, then $2^{n}-1$ is divisible by $2^{2}-1=3$. But the number $3^{n}-1$ is not divisible by 3. Therefore, we need to consider the case when $n=2 m+1$. Since $2^{4} \equiv 2^{2}(\bmod 12)$, then $2^{2 m} \equiv 4(\bmod 12)$, and hence, $2^{n}-1 \equiv 7(\bmod 12)$. Any prime number $p>3$ when divided by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,073 |
31.48. Prove that $g_{0}=0$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 31.48. It is clear that $g_{0}=\sum_{k=1}^{p-1}\left(\frac{k}{p}\right)$. Half of the numbers from 1 to $p-1$ are quadratic residues, and the rest are non-residues. | proof | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,074 |
31.49. Prove that $g_{1}^{2}=\left(\frac{-1}{p}\right) p$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 31.49. It is clear that
$$
g_{1}^{2}=\sum_{k, l=1}^{p-1}\left(\frac{k}{p}\right)\left(\frac{l}{p}\right) \varepsilon^{k+l}=\sum_{k, l=1}^{p-1}\left(\frac{k l}{p}\right) \varepsilon^{k+l}
$$
For a fixed $l$, the mapping $k \mapsto k l$ is a permutation of the residues modulo $p$, therefore
$$
\begin{aligned}
& \sum_{... | proof | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,075 |
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